## Calculating probability exciton is in a specific region

Anything regarding the post-processing utility (e.g. excitonic wavefunction analysis) is dealt with in this forum.

Moderators: Davide Sangalli, andrea marini, Daniele Varsano

claudio
Posts: 320
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### Re: Calculating probability exciton is in a specific region

Dear Alan

I replied to you too fast without reading all your message, sorry.

I wrote the electron/hole density in ypp, but without thinking to the spin-orbit case,

Probably it has to be generalized to the spin-orbit case, I will think about it, and let you know

best
Claudio
Claudio Attaccalite
[CNRS/ Aix-Marseille Université/ CINaM laborarory / TSN department
Campus de Luminy – Case 913
13288 MARSEILLE Cedex 09
web site: http://www.attaccalite.com
Freely download scientific books from: http://www.freescience.info

arb83@cam.ac.uk
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### Re: Calculating probability exciton is in a specific region

Dear Claudio

Many thanks for looking into this. In case it's of help I plotted the averaged electron and hole densities for the triplet state in the isolated organic molecule (without spin-orbit coupling) and they agree almost entirely with the averaged densities on the organic at the interface (with spin-orbit coupling - note the organic is just a hydrocarbon so should be unaffected by spin-orbit coupling). This also makes more physical sense within the system I'm studying. Therefore, I'm wondering if the issue lies with plotting a triplet state with fixed hole position at the interface when using spin-orbit coupling (I reproduce the triplet states well without spin-orbit coupling at the interface when fixing the hole).

I don't know how the spatial distributions are calculated with fixed hole position, but I'm wondering if this could be something to do with the nature of the triplet state (i.e. the spatial part is antisymmetric), plus some additional degrees of freedom due to spin-orbit coupling. I have found that higher energy states which aren't fully singlet or triplet agree fairly well between the averaging and fixed hole approaches.

All the best,

Alan
Alan Bowman
University of Cambridge

Davide Sangalli
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### Re: Calculating probability exciton is in a specific region

Dear all,
it is not easy to understand what is going on.

I just point to the lines of code where the excitonic WFs is computed.
The subroutine is

Code: Select all

``ypp/excitons/excitons_WFs.F``
I checked and there are two situations coded:

Code: Select all

``````   if(l_average_eh) then
...
else
...
endif
``````

In the first case (l_average_eh=.true.) I see indeed a bug if n_spinor=2.
This line

Code: Select all

``                v2plot(ir) = v2plot(ir)+B_coeff*WF1_ir(1)*conjg(WF2_ir(1))``
should be replaced by

Code: Select all

``````                v2plot(ir) = v2plot(ir)+B_coeff*WF1_ir(1)*conjg(WF2_ir(1))
if (n_spinor==2) then v2plot(ir) = v2plot(ir)+B_coeff*WF1_ir(2)*conjg(WF2_ir(2))
``````
and similarly for the electron

Hope it helps.
Best,
D.
Davide Sangalli, PhD
CNR-ISM, Division of Ultrafast Processes in Materials (FLASHit) and MaX Centre
http://www.oldism.cnr.it/en/davide-sangalli-cv/
http://www.max-centre.eu/

arb83@cam.ac.uk
Posts: 98
Joined: Thu Jul 02, 2020 3:56 pm

### Re: Calculating probability exciton is in a specific region

Dear Davide

Many thanks for this, I am now recompiling yambo, hopefully it works.

I have a more general question regarding what is being calculated here when spin-orbit coupling is included (and therefore how to interpret results). Without spin-orbit coupling I assume the exciton wavefunction is described as \Xsi=sum over states(\Psi_{electron state i}(r_{electron}) \Psi_{hole state i}(r_{hole})). Therefore, if you fix r_{hole} somewhere, you obtain a weighted sum over \Psi_{electron state i}, so we see something like where the electron will be. I assume this is what is done when fixing a hole somewhere using ypp (and similarly when calculating average positions, there is an integral over r_{hole}).

My confusion occurs when you include spin orbit coupling. I assume here the symmetry of the spatial wavefunction is included in the calculations, so it is not possible to separate the electron and hole position vectors, rather there is just r_{1} and r_{2}. Consider, for example, an antisymmetric state. At a guess (though probably more advanced in coding), this would be described by something like:

\Xsi=sum over states(\Psi_{electron state i}(r_{1}) \Psi_{hole state i}(r_{2}) - \Psi_{electron state i}(r_{2}) \Psi_{hole state i}(r_{1})).

In this case I'm not sure how you 'fix a hole' somewhere, or the physical meaning of this? What I'm concerned by is that, for example, if you look at the hole states and fix them at Rx, then (by reducing position vectors so there is only one) you would obtain

\Xsi=sum over states(\Psi_{electron state i}(r) \Psi_{hole state i}(Rx) - \Psi_{electron state i}(r) \Psi_{hole state i}(Rx))=0.

Within computation this wouldn't quite go to 0, and following normalisation it wouldn't be obvious this is what had happened.

I've outlined my question with fixing a hole somewhere, but I'd appreciate an answer also on the average position (where there is 'an integral over the hole position', which again I struggle with the physical meaning of). If this isn't clear I'm more than happy to give more details.

All the best,

Alan
Alan Bowman
University of Cambridge

arb83@cam.ac.uk
Posts: 98
Joined: Thu Jul 02, 2020 3:56 pm

### Re: Calculating probability exciton is in a specific region

Dear Davide

I realise that there is no 'endif' for the correction to the code you suggested above. Should I just put one on the following line or is there something else that needs to go in too?

All the best,

Alan
Alan Bowman
University of Cambridge

Daniele Varsano
Posts: 2719
Joined: Tue Mar 17, 2009 2:23 pm
Contact:

### Re: Calculating probability exciton is in a specific region

Dear Alan,

Code: Select all

``if (n_spinor==2) then v2plot(ir) = v2plot(ir)+B_coeff*WF1_ir(2)*conjg(WF2_ir(2))``
should be

Code: Select all

`````` v2plot(ir) = v2plot(ir)+B_coeff*WF1_ir(1)*conjg(WF2_ir(1))
if(n_spinor==2) v2plot(ir) = v2plot(ir)+B_coeff*WF1_ir(2)*conjg(WF2_ir(2))``````
Best,

Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/

arb83@cam.ac.uk
Posts: 98
Joined: Thu Jul 02, 2020 3:56 pm

### Re: Calculating probability exciton is in a specific region

Dear Daniele

Do you mean it should be:

v2plot(ir) = v2plot(ir)+B_coeff*WF1_ir(1)*conjg(WF2_ir(1))
if(n_spinor==2) then v2plot(ir) = v2plot(ir)+B_coeff*WF1_ir(2)*conjg(WF2_ir(2))

(i.e. with 'then' included)? Using the code I've been given ypp does not compile.

All the best,

Alan
Alan Bowman
University of Cambridge

Daniele Varsano
Posts: 2719
Joined: Tue Mar 17, 2009 2:23 pm
Contact:

### Re: Calculating probability exciton is in a specific region

Dear Alan,

see the two lines I wrote in the previous post:

Code: Select all

`````` v2plot(ir) = v2plot(ir)+B_coeff*WF1_ir(1)*conjg(WF2_ir(1))
if(n_spinor==2) v2plot(ir) = v2plot(ir)+B_coeff*WF1_ir(2)*conjg(WF2_ir(2))``````
"then" directive should be *not* included.

Best,
Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/

arb83@cam.ac.uk
Posts: 98
Joined: Thu Jul 02, 2020 3:56 pm

### Re: Calculating probability exciton is in a specific region

Dear Daniele

Many thanks, the code now complied!

I look forward to hearing back about my above question on what is being plotted/calculated when spin-orbit coupling is included, but until then all the best.

Best wishes,

Alan
Alan Bowman
University of Cambridge

arb83@cam.ac.uk
Posts: 98
Joined: Thu Jul 02, 2020 3:56 pm

### Re: Calculating probability exciton is in a specific region

Dear All

Would someone be able to comment on my post of 13th January please? If I am just missing something obvious and exciton/hole wavefunctions are still fully separable within the context of SO coupling please do just say (and any papers on the subject would be much appreciated!).

I hope you are all well.

With best wishes,

Alan
Alan Bowman
University of Cambridge